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Physics Solution Manual for 1100 and 2101

# Now 0 vv0 since the mass of mercury in the tube

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Unformatted text preview: ower temperatures. SOLUTION Substituting ∆V = βV0 ∆T (Equation 12.3) into Equation (1), we obtain V = V0 − β V0 ∆T or V = V0 (1 − β∆T ) Therefore, the volume V of the carbon tetrachloride at −13.0 °C is ){ ( ( )} V = 2.54 ×10−4 m3 1 − 1240 ×10−6 (Co )−1 75.0 o C − −13.0 o C = 2.26 ×10−4 m3 We have taken the value for β , the coefficient of volume expansion for carbon tetrachloride, from Table 12.1 in the text. 31. SSM REASONING The increase ∆V in volume is given by Equation 12.3 as ∆V = βV0∆T, where β is the coefficient of volume expansion, V0 is the initial volume, and ∆T is the increase in temperature. The lead and quartz objects experience the same change in volume. Therefore, we can use Equation 12.3 to express the two volume changes and set them equal. We will solve the resulting equation for ∆TQuartz. SOLUTION Recognizing that the lead and quartz objects experience the same change in volume and expressing that change with Equation 12.3, we have β LeadV0 ∆TLead = β QuartzV0 ∆TQuartz 14 244 4 3 1442443 4 4 ∆VLead ∆V Quartz In this result V0 is the initial volume of each object. Solving for ∆TQuartz and taking values for the coefficients of volume expansion for lead and quartz from Table 12.1 gives −1 ∆TQuartz −6 β Lead ∆TLead 87 ×10 ( C° ) ( 4.0 C° ) = = = 230 C° −1 β Quartz 1.5 ×10−6 ( C° ) Chapter 12 Problems 645 32. REASONING The density ρ of a substance is defined as its mass m divided by its volume V0 (Equation 11.1); ρ = m/V0. When the temperature rises, the volume increases by an amount ∆V = β V0 ∆T (Equation 12.3), where β is the coefficient of volume expansion and ∆T is the change in temperature. Since the mass of the fluid does not change as the temperature rises, the increase in volume means that the density decreases. SOLUTION a. The density of the fluid at 0 °C is ρ= m 825 kg = = 705 kg/m3 V0 1.17 m3 b. When the temperature rises to 20.0 °C from 0 °C the volume increases to V0 + ∆V, where ∆V = β V0 ∆T . Therefore, the density becomes ρ= = m m m = = V0 + ∆V V0 + β V0 ∆T V0 (1 + β ∆T ) 825 kg (1.17 m3 ) {1 + 1.26 ×10−3 ( C° ) } −1 = 688 kg/m3 ( 20.0 °C − 0 °C ) ______________________________________________________________________________ 33. REASONING AND SOLUTION Both the coffee and beaker expand as the temperature increases. For the expansion of the coffee (water) ∆VW = βWV0 ∆T For the expansion of the beaker (Pyrex glass) ∆VG = βG V0 ∆T The excess expansion of the coffee, hence the amount which spills, is ∆V = ∆VW − ∆VG = (βW − βG) V0 ∆T Taking the coefficients of volumetric expansion βW and coffee (water) and βG for the beaker (Pyrex glass) from Table 12.1, we find ∆V = ( β W − βG )V0 ∆T ( ) –1 –1 = 207 ×10 –6 ( C° ) − 9.9 × 10 –6 ( C° ) 0.50 ×10 –3 m3 ( 92 °C –18 °C ) = 7.3 × 10 –6 m3 646 TEMPERATURE AND HEAT ______________________________________________________________________________ 34. REASONING According to ∆V = β V0 ∆T (Equation 12.3), the change ∆V in volume depends on the coefficient of volume expansion β, the initial volume V0, and the change ∆T in temperature. The liquid expands more because its coefficient of volume expansion is larger, and the change in volume is directly proportional to that coefficient. The volume of liquid that spills over is equal to the change in volume of the liquid minus the change in volume of the can. SOLUTION The volume of liquid that spills over the can is the difference between the increase in the volume of the liquid and that of the aluminum can: ∆V = β LV0 ∆T − β AV0 ∆T Therefore, βL = = ∆V + βA V0 ∆T −6 3.6 × 10 m (3.5 × 10 −4 m 3 3 ) ( 78 °C − 5 °C ) + 69 × 10−6 ( C° ) −1 = 2.1 × 10−4 ( C° ) −1 ______________________________________________________________________________ 35. SSM REASONING AND SOLUTION Both the gasoline and the tank expand as the temperature increases. The coefficients of volumetric expansion β g and βs for gasoline and steel are available in Table 12.1. According to Equation 12.3, the volume expansion of the gasoline is −1 ∆Vg = β gV0 ∆T = 950 × 10−6 ( C° ) (20.0 gal)(18 C°) = 0.34 gal while the volume of the steel tank expands by an amount −1 ∆Vs = β sV0 ∆T = 36 × 10−6 ( C° ) (20.0 gal)(18 C°) = 0.013 gal The amount of gasoline which spills out is ∆Vg − ∆Vs = 0.33 gal ______________________________________________________________________________ Chapter 12 Problems 647 36. REASONING AND SOLUTION Both the water and pipe expand as the temperature increases. For the expansion of the water ∆Vw = βw V0 ∆T For the expansion of the copper pipe ∆Vc = βc V0 ∆T The initial volume of the pipe and water is V0 = π r2L The reservoir then needs a capacity of ∆V = ∆Vw − ∆Vc = (β w − β c) V0 ∆T Taking the coefficients of volumetric expansion βw and β c for water and copper from Table 12.1,...
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